Question 1208809
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For any solution we find, changing the sign of any one of the numbers gives another solution, because (-a)^2 is equal to a^2.  So to start with we only need to look for non-negative solutions.<br>
The problem does not specify positive integers, so 0 is a possibility.  And since 169 = 13^2, the triple of numbers (13,0,0) is a solution.<br>
Those three numbers can be in any order -- i.e., (0,13,0) and (0,0,13) are also solutions.  So there are 3 solutions using the non-negative integers 13, 0, and 0.<br>
But any of those numbers can be replace by its opposite.  The "opposite" of 0 is still 0, so we can't change the sign of any 0 to get a new solution.  But we can replace each 13 with -13 to get a new solution.<br>
Summary of the problem so far:
13, 0, and 0 satisfy the equation (1 solution);
there are 3 ways to arrange those three numbers (1*3 = 3 solutions);
we can change the sign of the 13 (3*2 = 6 solutions).<br>
So there are 6 solutions using the number 13, 0, 0, and their opposites.<br>
Now look for other solutions.  To do that, use a "greedy" algorithm -- that is, try the largest numbers first.  We have used 13, so now look at 12.<br>
12^2 = 144; 169-144 = 25; and that can be written as 16+9 = 4^2+3^2.<br>
So the numbers 12, 4, and 3 provide another solution -- or, in fact, a large set of solutions.<br>
Rearranging the three numbers and changing any of their signs, similar to what we did with the three number 13, 0, and 0, we get the following summary:
12, 4, and 3 satisfy the equation (1 solution);
there are 6 ways to arrange those three numbers (1*6 = 6 solutions);
we can change the sign, or not change the sign, of EACH of the three numbers (6*2*2*2 = 48 solutions).<br>
Continuing with our greedy algorithm, we quickly find that there is no other set of three numbers which will give a solution to the given equation.<br>
So...<br>
ANSWER: There are 6+48 = 54 solutions to the given equation<br>
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Thanks to tutor @ikleyn for pointing out the missing solutions in my original answer.<br>
I used 0 in the set 13, 0, and 0; so I should have remembered to include the set 12, 5, and 0.<br>
The number of missing solutions in my original solution is...<br>
12, 5, 0: (1 solution)
arranging those in any of 6 ways: 1*6 = 6 solutions
changing the sign of either or both of the 12 and 5: 6*2*2 = 24 solutions<br>
Final (corrected) answer: 54+24 = 78<br>